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Elden [556K]
3 years ago
14

1. Vector B is multiplied by the scalar 921.5. What angle does the new vector make with the x-axis?

Physics
2 answers:
Ivanshal [37]3 years ago
7 0

Answer:

#1 is actually 33 because I’m assuming the angle doesn’t change just because the vector line gets longer

Explanation: I took this question on physics test and picked 49.5 and I got it wrong

adelina 88 [10]3 years ago
6 0
1. Is 49.5
2. Is 8.6
4. Is 6.6
Wait I’m not sure
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If you were standing at the center of curvature in front of a concave mirror, what image would be projected?
matrenka [14]
"<span>The image would be upside down, would look as tall as you, and would be at the same distance from the mirror as you are" is the type of image among the choices given in the question that would be projected. The correct option among all the options that are given in the question is the first option. I hope it helps you.</span>
6 0
3 years ago
Read 2 more answers
A boy with a mouse is 80 kg stands on a scale in an ascending elevator with an acceleration of 3 m/s What is the resultant upwar
Rus_ich [418]

Answer:

Upward force on the boy will be 1024 N      

Explanation:

We have given mass of the boy m = 80 kg

Acceleration due to gravity g=9.8m/sec^2

It is given that elevator is ascending with acceleration of 3m/sec^2

So net acceleration on the boy = 9.8+3 = 12.8 m/sec^2 ( As the car elevator is moving ascending )

So upward force on the boy will be equal to F = m(g+a)

So F=80\times 12.8=1024N

So upward force on the boy will be 1024 N

3 0
3 years ago
What is the net force acting on the piano
CaHeK987 [17]

answer: 500 net force

5 0
3 years ago
Assume that the driver begins to brake the car when the distance to the wall is d=107m, and take the car's mass as m-1400kg, its
Evgen [1.6K]

Answer:

Explanation:

a ) Let let the frictional force needed be F

Work done by frictional force = kinetic energy of car

F x 107 = 1/2 x 1400 x 35²

F = 8014 N

b )

maximum possible static friction

= μ mg

where μ is coefficient of static friction

= .5 x 1400 x 9.8

= 6860 N

c )

work done by friction for μ = .4

= .4 x 1400 x 9.8 x 107

= 587216 J

Initial Kinetic energy

= .5 x 1400 x 35 x 35

= 857500 J

Kinetic energy at the at of collision

= 857500 - 587216

= 270284 J

So , if v be the velocity at the time of collision

1/2 mv² = 270284

v = 19.65 m /s

d ) centripetal force required

= mv₀² / d which will be provided by frictional force

= (1400 x 35 x 35) / 107

= 16028 N

Maximum frictional force possible

= μmg

= .5 x 1400 x 9.8

= 6860 N

So this is not possible.

4 0
3 years ago
In an aqueous solution where the H+ concentration is 1 x 10-6 M, the OH concentration must be:
vesna_86 [32]

Answer:

D. 1×10⁻⁸ M

Explanation:

[H⁺] [OH⁻] = 10⁻¹⁴

(1×10⁻⁶) [OH⁻] = 10⁻¹⁴

[OH⁻] = 1×10⁻⁸

6 0
3 years ago
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